Let $G$ be a finite group, and let $N \lhd G$ such that $C_G(N) \leq N$.
Suppose that $N=A \times B$, that $A \lhd G$ and $B \lhd G$, and that moreover $\operatorname{Out}(A)=1$. I am tempted to say that this implies that $G= A \times C$ for some normal subgroup $C \lhd G$, but I can't prove it directly.
Of course using the NC theorem $G/N$ is a subgroup of $\operatorname{Out}(N)$, and hence of $\operatorname{Out}(B)$, but is this enough to say that $A$ is a direct factor of $G$?
The group of order $12$ with presentation $\langle x,y \mid x^4=y^3=1,x^{-1}yx=y^{-1} \rangle$ (this is sometimes called the dicyclic group of order $12$) is a counterexample with $A = \langle x^2 \rangle$, $B = \langle y \rangle$.